6. Tom who is 5 feet tall threw an object into the air. The quadratic function f(x)= - 16x^2 + 64x +5 is where f(x) is the height of the object in feet and x is the time in seconds. The function in vertex form is
a) f(x) = -16(x-2)^2 -11
b) f(x) = -16(x-2)^2 - 59
c) f(x) = -16(x-2)^2 + 5
d) f(x) = -16 (x-2)^2 + 69
7. The object hits the ground after about ____ seconds. Choose the correct answer from the following a) 2 b) 3 c) 4 d) 5
To put the quadratic function f(x) = -16x^2 + 64x + 5 in vertex form, we need to identify the vertex of the parabola.
The vertex form of a quadratic function is f(x) = a(x-h)^2 + k, where (h, k) represents the vertex of the parabola.
The x-coordinate of the vertex is found using the formula x = -b/2a. In this case, a = -16 and b = 64.
x = -64/2(-16)
x = 64/32
x = 2
To find the y-coordinate of the vertex, substitute the value of x = 2 into the original function:
f(2) = -16(2)^2 + 64(2) + 5
f(2) = -16(4) + 128 + 5
f(2) = -64 + 128 + 5
f(2) = 69
Therefore, the vertex of the parabola is (2, 69).
Thus, the function in vertex form is f(x) = -16(x-2)^2 + 69.
The object hits the ground when f(x) = 0, so we need to solve the quadratic equation -16x^2 + 64x + 5 = 0.
Using the quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a), we find the roots of the equation:
x = (-64 ± √(64^2 - 4(-16)(5)))/(2(-16))
x = (-64 ± √(4096 + 320))/(2(-16))
x = (-64 ± √4416)/-32
x = (-64 ± 66.51)/-32
The positive root is the time it takes for the object to hit the ground after being thrown into the air. Taking the positive root:
x ≈ (-64 + 66.51)/-32
x ≈ 2.51/-32
x ≈ -0.078
Since the time in seconds cannot be negative, the object hits the ground after about 0.078 seconds.
Therefore, the correct answers are:
6. The function in vertex form is d) f(x) = -16 (x-2)^2 + 69.
7. The object hits the ground after about a) 2 seconds.